37. If DS° for the following reaction is -149 J/mol. Calculate S° for Cl 2 (g).

Question

37. If DS° for the following reaction is -149 J/mol. Calculate S° for Cl2 (g).

            2Fe(s) + 3Cl2 (g) à 2FeCl3(s)

Standard entropies for Fe = 27 J/mol; FeCl3 = 142 J/mol

a)

378.9 J/mol

b)

126.3 J/mol

c)

264 J/mol

d)

88 J/mol

38. What is the highest temperature at which the following reaction becomes spontaneous?

A (aq) + B (aq) C (s) + D (aq)       

H°= -34.0 kJ/mol                  S° = -57.0 J/mol•K

a). 130°C                     b). 323°C                    c). 552°C                     d). 705°C        e). At all temperatures.

39. What is the value of G° (in kJ/mol) for this reaction at 25°C?

A (g) B (g)

H° = -42.4 kJ/mol S°= -56 J/mol•K

a. -136                            b. +236                     c. -26               d. -59               e. none of the above

a)

378.9 J/mol

b)

126.3 J/mol

c)

264 J/mol

d)

88 J/mol

Explanation / Answer

37

2Fe(s) + 3Cl2 (g) ---> 2FeCl3(s)

for

S° for Cl2 (g).

dS = 2S-FeCl3 - (3S-Cl2 + 2S-Fe)

dS = 2*142-(3*Cl2 + 2*27) = -149

S-Cl2 =  (-149 - 2*142+2*27))/(-3) =126.3333 J/mol

answer is B

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